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Please help me to solve the differential equation $$y''=\frac{1}{y^2}$$

Thank you very much!

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Try a solution of the form $y=at^b$, where $a$ and $b$ are constants. Use the diffeq to find $a$ and $b$. The solution can then be generalized by the substitution $t\rightarrow t+c$. – Ben Crowell Oct 3 '12 at 15:20

A start: The following is a standard trick. Multiply both sides by $y'$. Then the new left side is the derivative of $(1/2)(y')^2$ and the new right side is the derivative of $-1/y$. It follows that $$\frac{1}{2}(y')^2= -\frac{1}{y}+C$$ for some constant $C$. Solve for $y'$. You will get a couple of separable first-order differential equations.

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Physically, the diffeq can be interpreted as Newton's second law for a charged particle in the field of a second, like charge fixed at y=0. The method in your answer can be interpreted as a statement of conservation of energy. – Ben Crowell Oct 3 '12 at 15:17

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